Determine the equations of motion for the two masses described in Problem \({\bf{1}}\)if\({{\bf{m}}_{\bf{1}}}{\bf{ = 1\;kg,}}{{\bf{m}}_{\bf{2}}}{\bf{ = 1\;kg}}\),\({{\bf{k}}_{\bf{1}}}{\bf{ = 3\;N/m}}\), and\({{\bf{k}}_{\bf{2}}}{\bf{ = 2\;N/m}}\).

Assume that \(x > 0,{\bf{ }}y > 0\) and \(y > x\) to derive the differential equations of motions, and they must be true for any x and y. The Forces are proportional to the change of the length of the string, so one has\({F_1} = {k_1}\left( {y - x} \right),\;{F_2} = - {k_1}\left( {y - x} \right),\;{F_3} = - {k_2}y\)

From the second Newton's law one now has:

\(\begin{aligned}{l}{m_1}x'' = {k_1}\left( {y - x} \right)\\{m_2}y'' = - {k_1}\left( {y - x} \right) - {k_2}y\end{aligned}\)

In our case\({m_1} = 1\;kg,{m_2} = 1\;kg,{k_1} = 3\;N/m,{k_2} = 2\;N/m\), so the system is:

\(\begin{aligned}{c}x'' &= 3(y - x) \Leftrightarrow x'' + 3x - 3y = 0\\y'' &= - 3(y - x) - 2y \Leftrightarrow - 3x + y'' + 5y &= 0\end{aligned}\)

One will solve this system using the elimination method. First,one needs to rewrite this system in operator form:

\(\begin{aligned}{c}\left( {{D^2} + 3} \right){\rm{(}}x{\rm{)}} - 3{\rm{(}}y{\rm{)}} = 0\\ - 3{\rm{(}}x{\rm{)}} + \left( {{D^2} + 5} \right){\rm{(}}y{\rm{)}} &= 0\end{aligned}\)

Multiplying the first equation by \(\left( {{D^2} + 5} \right)\) and the second by 3, and then adding them together give us

\(\begin{aligned}{c}\left( {{D^2} + 5} \right)\left( {{D^2} + 3} \right){\rm{(}}x{\rm{)}} - 3\left( {{D^2} + 5} \right){\rm{(}}y{\rm{)}} &= 0\\ - 9{\rm{(}}x{\rm{)}} + 3\left( {{D^2} + 5} \right){\rm{(}}y{\rm{)}} &= 0\\\left( {\left( {{D^2} + 5} \right)\left( {{D^2} + 3} \right) - 9} \right){\rm{(}}x{\rm{)}} &= 0 \Leftrightarrow \left( {{D^4} + 8{D^2} + 6} \right){\rm{(}}x{\rm{)}} = 0\end{aligned}\)

The auxiliary equation is\({r^4} + 8{r^2} + 6 = 0\), and its solutions are:

\(\begin{aligned}{c}{r^2} = \frac{{ - 8 \pm \sqrt {68 - 24} }}{2} &= - 4 \pm \sqrt {10} \\{r_{1,2}} &= \pm i\sqrt {4 - \sqrt {10} } ,\;\;\;{r_{3,4}} &= \pm i\sqrt {4 + \sqrt {10} } \end{aligned}\)

So, the general solution for x is:

\(x\left( t \right) = {c_1}\cos \left( {\sqrt {4 - \sqrt {10} } t} \right) + {c_2}\sin \left( {\sqrt {4 - \sqrt {10} } t} \right) + {c_3}\cos \left( {\sqrt {4 + \sqrt {10} } t} \right) + {c_4}\sin \left( {\sqrt 4 + \sqrt {10} t} \right)\)

From the first equation of the system, one has that\(3y = x'' + 3x\), so let’s first find\({D^2}{\rm{(}}x{\rm{)}}\):

\(\begin{aligned}{c}D{\rm{(}}x{\rm{)}} &= - \sqrt {4 - \sqrt {10} } {c_1}\sin \left( {\sqrt {4 - \sqrt {10} } t} \right) + \sqrt {4 - \sqrt {10} } {c_2}\cos \left( {\sqrt {4 - \sqrt {10} } t} \right) - \sqrt {4 + \sqrt {10} } {c_3}\sin \left( {\sqrt {4 + \sqrt {10} } t} \right) + \sqrt {4 + \sqrt {10} } {c_4}\cos \left( {\sqrt {4 + \sqrt {10} } t} \right)\\{D^2}{\rm{(}}x{\rm{)}} &= - \left( {4 - \sqrt {10} } \right){c_1}\cos \left( {\sqrt {4 - \sqrt {10} } t} \right) - \left( {4 - \sqrt {10} } \right){c_2}\sin \left( {\sqrt {4 - \sqrt {10} } t} \right) - (4 + \sqrt {10} ){c_3}\cos \left( {\sqrt {4 + \sqrt {10} } t} \right) - \left( {4 + \sqrt {10} } \right){c_4}\sin \left( {\sqrt {4 + \sqrt {10} } t} \right)\end{aligned}\)

Now one has,

\(\begin{aligned}{c}3y\left( t \right) &= - \left( {4 - \sqrt {10} } \right){c_1}\cos \left( {\sqrt {4 - \sqrt {10} } t} \right) - \left( {4 - \sqrt {10} } \right){c_2}\sin \left( {\sqrt {4 - \sqrt {10} } t} \right) - \left( {4 + \sqrt {10} } \right){c_3}\cos \left( {\sqrt {4 + \sqrt {10} } t} \right) - \left( {4 + \sqrt {10} } \right){c_4}\sin \left( {\sqrt {4 + \sqrt {10} } t} \right)\\ + 3{c_1}\cos \left( {\sqrt {4 - \sqrt {10} } t} \right) + 3{c_2}\sin \left( {\sqrt {4 - \sqrt {10} } t} \right) + 3{c_3}\cos \left( {\sqrt {4 + \sqrt {10} } t} \right) + 3{c_4}\sin \left( {\sqrt {4 + \sqrt {10} } t} \right)\\ &= - \left( {1 - \sqrt {10} } \right){c_1}\cos \left( {\sqrt {4 - \sqrt {10} } t} \right) - \left( {1 - \sqrt {10} } \right){c_2}\sin \left( {\sqrt {4 - \sqrt {10} } t} \right) - \left( {1 + \sqrt {10} } \right){c_3}\cos \left( {\sqrt {4 + \sqrt {10} } t} \right) - (1 + \sqrt {10} ){c_4}\sin \left( {\sqrt {4 + \sqrt {10} } t} \right)\\y\left( t \right) &= - \frac{{1 - \sqrt {10} }}{3}{c_1}\cos \left( {\sqrt {4 - \sqrt {10} } t} \right) - \frac{{1 - \sqrt {10} }}{3}{c_2}\sin \left( {\sqrt {4 - \sqrt {10} } t} \right) - \frac{{1 + \sqrt {10} }}{3}{c_3}\cos \left( {\sqrt {4 + \sqrt {10} } t} \right) - \frac{{1 + \sqrt {10} }}{3}{c_4}\sin \left( {\sqrt {4 + \sqrt {10} } t} \right)\end{aligned}\)

The initial conditions are\(x\left( 0 \right) = - 1,x'\left( 0 \right) = y\left( 0 \right) = y'\left( 0 \right) = 0\).

First, one needs to find the first derivative of\({\bf{y}}\):

\(\begin{aligned}{c}D{\rm{(}}y{\rm{)}} &= \frac{{\left( {1 - \sqrt {10} } \right)\sqrt {4 - \sqrt {10} } }}{3}{c_1}\sin \left( {\sqrt {4 - \sqrt {10} } t} \right) - \frac{{\left( {1 - \sqrt {10} } \right)\sqrt {4 - \sqrt {10} } }}{3}{c_2}\cos \left( {\sqrt {4 - \sqrt {10} } t} \right){\bf{ }}\\ + \frac{{\left( {1 + \sqrt {10} } \right)\sqrt {4 + \sqrt {10} } }}{3}{c_3}\sin \left( {\sqrt {4 + \sqrt {10} } t} \right) - \frac{{\left( {1 + \sqrt {10} } \right)\sqrt {4 + \sqrt {10} } }}{3}{c_4}\sin \left( {\sqrt {4 + \sqrt {10} } t} \right)\end{aligned}\)

So, the initial conditions give us:\(x\left( 0 \right) = {c_1} + {c_3} = - 1,\)

\(\begin{aligned}{c}x'\left( 0 \right) &= \sqrt {4 - \sqrt {10} } {c_2} + \sqrt {4 + \sqrt {10} } {c_4} &= 0\\y\left( 0 \right) &= - \frac{{1 - \sqrt {10} }}{3}{c_1} - \frac{{1 + \sqrt {10} }}{3}{c_3} &= 0\\y'\left( 0 \right) &= - \frac{{\left( {1 - \sqrt {10} } \right)\sqrt {4 - \sqrt {10} } }}{3}{c_2} - \frac{{\left( {1 + \sqrt {10} } \right)\sqrt {4 + \sqrt {10} } }}{3}{c_4} &= 0\end{aligned}\)

From the second equation one has that\({c_2} = \frac{{ - \sqrt {4 + \sqrt {10} } }}{{\sqrt {4 - \sqrt {10} } }}{c_4}\), so substituting this into the fourth equation one has:

\(\begin{aligned}{c} - \frac{{\left( {1 - \sqrt {10} } \right)\sqrt {4 - \sqrt {10} } }}{3} \times \left( { - \frac{{4 + \sqrt {10} }}{{4 - \sqrt {10} }}} \right){c_4} - \frac{{\left( {1 + \sqrt {10} } \right)\sqrt {4 + \sqrt {10} } }}{3}{c_4} &= 0\\ \Leftrightarrow \left( {1 - \sqrt {10} + 1 + \sqrt {10} } \right){c_4} = 0\\ \Leftrightarrow {c_4} = 0\; \Rightarrow {c_2} = 0\end{aligned}\)

Multiplying the third equation \(y\left( 0 \right) = 0\) by 3 and expressing \({c_1}\) in terms of \({c_3}\) gives us:

\({c_1} = \frac{{ - \left( {1 + \sqrt {10} } \right)}}{{\left( {1 - \sqrt {10} } \right){c_3}}}\)

Substituting this into the first equation we get\( - \frac{{1 + \sqrt {10} }}{{1 - \sqrt {10} }}{c_3} + {c_3} = - 1\)

\(\begin{aligned}{c} \Leftrightarrow \left( { - 1 - \sqrt {10} + 1 - \sqrt {10} } \right){c_3} &= - 1 \times \left( {1 - \sqrt {10} } \right)\\ \Leftrightarrow {c_3} &= \frac{{1 - \sqrt {10} }}{{2\sqrt {10} }} &= \frac{{\sqrt {10} - 10}}{{20}}\\ \Rightarrow {c_1} &= - \frac{{1 + \sqrt {10} }}{{1 - \sqrt {10} }} \times \frac{{1 - \sqrt {10} }}{{2\sqrt {10} }} &= - \frac{{\sqrt {10} + 10}}{{20}}\end{aligned}\)

Substituting values for \({c_1},{\bf{ }}{c_2},{\bf{ }}{c_3}\) and\({c_4}\) into the general solutions for x and y one has:

\(\begin{aligned}{l}{\bf{x(t) = - }}\frac{{\sqrt {{\bf{10}}} {\bf{ + 10}}}}{{{\bf{20}}}}{\bf{cos(}}\sqrt {{\bf{4 - }}\sqrt {{\bf{10}}} } {\bf{t) + }}\frac{{\sqrt {{\bf{10}}} {\bf{ - 10}}}}{{{\bf{20}}}}{\bf{cos(}}\sqrt {{\bf{4 + }}\sqrt {{\bf{10}}} } {\bf{t),}}\\{\bf{y(t) = - }}\frac{{{\bf{1 - }}\sqrt {{\bf{10}}} }}{{\bf{3}}}{{\bf{c}}_{\bf{1}}}{\bf{cos(}}\sqrt {{\bf{4 - }}\sqrt {{\bf{10}}} } {\bf{t) - }}\frac{{{\bf{1 + }}\sqrt {{\bf{10}}} }}{{\bf{3}}}{{\bf{c}}_{\bf{3}}}{\bf{cos(}}\sqrt {{\bf{4 + }}\sqrt {{\bf{10}}} } {\bf{t)}}\\{\bf{y(t) = - }}\frac{{{\bf{3}}\sqrt {{\bf{10}}} }}{{{\bf{20}}}}{\bf{cos(}}\sqrt {{\bf{4 - }}\sqrt {{\bf{10}}} } {\bf{t) + }}\frac{{{\bf{3}}\sqrt {{\bf{10}}} }}{{{\bf{20}}}}{\bf{cos(}}\sqrt {{\bf{4 + }}\sqrt {{\bf{10}}} } {\bf{t)}}\end{aligned}\)